工业混合碱各组分含量测定——一种综合型实验方案设计

王园朝, 程和勇

Design of a Comprehensive Experiment for Determination of Each Component in Industrial Mixed Alkali

Yuanchao Wang, Heyong Cheng

表1 4种不同滴定方法测定两类混合碱组分质量浓度的计算公式

混合碱I (Na2CO3 + NaOH) 混合碱II (Na2CO3 + NaHCO3) 混合样品总体积/mL 25.00 25.00 方法一 酚酞为指示剂消耗HCl平均体积/mL V1 V3 甲基橙为指示剂消耗HCl平均体积/mL V2 V4 组分质量浓度ρ/(g∙L−1) ${\rho _{NaOH}} = \frac{{({V_1} - {V_2}){c_{{\text{HCl}}}} \times {M_{{\text{NaOH}}}}}}{{25.00\;}}$ ${\rho _{{\text{N}}{{\text{a}}_{\text{2}}}{\text{C}}{{\text{O}}_{\text{3}}}}} = \frac{{2{V_3}{c_{{\text{HCl}}}} \times \frac{{{M_{{\text{N}}{{\text{a}}_{\text{2}}}{\text{C}}{{\text{O}}_{\text{3}}}}}}}{2}}}{{25.00\;}}$ ${\rho _{{\text{N}}{{\text{a}}_{\text{2}}}{\text{C}}{{\text{O}}_{\text{3}}}}} = \frac{{2{V_2}{c_{{\text{HCl}}}} \times \frac{{{M_{{\text{N}}{{\text{a}}_{\text{2}}}{\text{C}}{{\text{O}}_{\text{3}}}}}}}{2}}}{{25.00\;}}$ ${\rho _{{\text{NaHC}}{{\text{O}}_{\text{3}}}}} = \frac{{({V_4} - {V_3}){c_{{\text{HCl}}}} \times {M_{{\text{NaHC}}{{\text{O}}_{\text{3}}}}}}}{{25.00\;}}$ 方法二 酚酞为指示剂消耗HCl平均体积/mL V5 V7 甲基橙为指示剂消耗HCl平均体积/mL V6 V8 组分质量浓度ρ/(g∙L−1) ${\rho _{{\text{NaOH}}}} = \frac{{{\text{(2}}{V_5} - {V_6}){c_{{\text{HCl}}}} \times {M_{{\text{NaOH}}}}}}{{25.00\;}}$ ${\rho _{{\text{N}}{{\text{a}}_{\text{2}}}{\text{C}}{{\text{O}}_{\text{3}}}}} = \frac{{2{V_7}{c_{{\text{HCl}}}} \times \frac{{{M_{{\text{N}}{{\text{a}}_{\text{2}}}{\text{C}}{{\text{O}}_{\text{3}}}}}}}{2}}}{{25.00\;}}$ ${\rho _{{\text{N}}{{\text{a}}_{\text{2}}}{\text{C}}{{\text{O}}_{\text{3}}}}} = \frac{{2({V_6} - {V_5}){c_{{\text{HCl}}}} \times \frac{{{M_{{\text{N}}{{\text{a}}_{\text{2}}}{\text{C}}{{\text{O}}_{\text{3}}}}}}}{2}}}{{25.00\;}}$ ${\rho _{{\text{NaHC}}{{\text{O}}_{\text{3}}}}} = \frac{{({V_8} - 2{V_7}){c_{{\text{HCl}}}} \times {M_{{\text{NaHC}}{{\text{O}}_{\text{3}}}}}}}{{25.00\;}}$ 方法三 酚酞为指示剂消耗HCl平均体积/mL V9 – 甲基橙为指示剂消耗HCl平均体积/mL V10 – 组分质量浓度ρ/(g∙L−1) ${\rho _{{\text{NaOH}}}} = \frac{{{V_9}{c_{{\text{HCl}}}} \times {M_{{\text{NaOH}}}}}}{{25.00\;}}$ – ${\rho _{{\text{N}}{{\text{a}}_{\text{2}}}{\text{C}}{{\text{O}}_{\text{3}}}}} = \frac{{({V_{10}} - {V_9}){c_{{\text{HCl}}}} \times \frac{{{M_{{\text{N}}{{\text{a}}_{\text{2}}}{\text{C}}{{\text{O}}_{\text{3}}}}}}}{2}}}{{25.00\;{\kern 1pt} }}$ – 方法四 第一化学计量点消耗HCl平均体积/mL V11 V13 第二化学计量点消耗HCl平均体积/mL V12 V14 组分质量浓度ρ/(g∙L−1) ${\rho _{{\text{NaOH}}}} = \frac{{(2{V_{11}} - {V_{12}}){c_{{\text{HCl}}}} \times {M_{{\text{NaOH}}}}}}{{25.00{\kern 1pt} \;}}$ ${\rho _{{\text{N}}{{\text{a}}_{\text{2}}}{\text{C}}{{\text{O}}_{\text{3}}}}} = \frac{{2{V_{13}}{c_{{\text{HCl}}}} \times \frac{{{M_{{\text{N}}{{\text{a}}_{\text{2}}}{\text{C}}{{\text{O}}_{\text{3}}}}}}}{2}}}{{25.00\;}}$ ${\rho _{{\text{N}}{{\text{a}}_{\text{2}}}{\text{C}}{{\text{O}}_{\text{3}}}}} = \frac{{2({V_{12}} - {V_{11}}){c_{{\text{HCl}}}} \times \frac{{{M_{{\text{N}}{{\text{a}}_{\text{2}}}{\text{C}}{{\text{O}}_{\text{3}}}}}}}{2}}}{{25.00\;}}$ ${\rho _{{\text{NaHC}}{{\text{O}}_{\text{3}}}}} = \frac{{({V_{14}} - 2{V_{13}}){c_{{\text{HCl}}}} \times {M_{{\text{NaHC}}{{\text{O}}_{\text{3}}}}}}}{{25.00\;}}$